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Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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26 Jul 2014, 00:11

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hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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29 Jul 2014, 19:55

2

This post received KUDOS

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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30 Jul 2014, 03:41

hubahuba wrote:

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.

You are Welcome. _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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18 Aug 2014, 12:48

VenoMftw wrote:

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

When you reach this point, how do you know that solving will give vales that won't be integers? Did you actually try to solve or do you have a shortcut?

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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19 Aug 2014, 16:34

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

I couldn't understand this line k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values. Where does the question say that the sum and product of the roots are integers.
_________________

I couldn't understand this line k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values. Where does the question say that the sum and product of the roots are integers.

It is not given that sum and product need to be integers. The logic of this question is a little more twisted:

Question: Are both roots positive?

Answer: Yes/No Or May be If we answer with a certain Yes or a certain No, our stmnt is sufficient. If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.

2x^2 + hx + k = 0 Sum of roots = -h/2 Product of roots = k/2

Let's find all those cases in which we can answer with 'Yes'

Stmnt 1: The product of the roots is 6. k/2 = 6 so k = 12

The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0 The roots could be no integer for any random value of h. So we have no definite answer from this statement. Not sufficient.

Stmnt 2: kh = 132 Let's see whether given this information, we can find both integer roots. kh = 132 = 2^2 * 3 * 11 If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers) So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign): k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No. k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No. So it is not possible to have both integers. We can answer with a definite 'No'.

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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23 Aug 2015, 21:25

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Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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27 Aug 2016, 02:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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If h and k are both positive integers, are both roots of the equation [#permalink]

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27 Oct 2016, 05:36

Hey, basically I came to the correct answer B but in other way:

(1) x1*x2=6=k/2 k=12 2x^2+hx+12=0

Basicaly, roots might be integers or non-integers so insufficient.

(2) h*k=132 (I) x1+x2=-h/2 (II) x1*x2=k/2 (III)

then I multiply (II) * (III) and get x1*x2*(x1+x2)=-k*h/4=-132/4=-33

So, if we asssume that x1 and x2 integers, what we should conclude is INTEGER1*INTEGER2*INTEGER3=-33 (which is possible only for triplet 1/3/11 with one/all of them negative - clearly among 1/3/11 there are no pairs to satisfy a+b=c =>> therefore, x1 OR x2 should be non-integer). So (2) is sufficient.

Not sure is it correct reasoning, but the answer is OK.

If h and k are both positive integers, are both roots of the equation [#permalink]

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31 Oct 2016, 23:21

VeritasPrepKarishma wrote:

It is not given that sum and product need to be integers. The logic of this question is a little more twisted:

Question: Are both roots positive?

Answer: Yes/No Or May be If we answer with a certain Yes or a certain No, our stmnt is sufficient. If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.

2x^2 + hx + k = 0 Sum of roots = -h/2 Product of roots = k/2

Let's find all those cases in which we can answer with 'Yes'

Stmnt 1: The product of the roots is 6. k/2 = 6 so k = 12

The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0 The roots could be no integer for any random value of h. So we have no definite answer from this statement. Not sufficient.

Stmnt 2: kh = 132 Let's see whether given this information, we can find both integer roots. kh = 132 = 2^2 * 3 * 11 If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers) So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign): k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No. k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No. So it is not possible to have both integers. We can answer with a definite 'No'.

It is not given that sum and product need to be integers. The logic of this question is a little more twisted:

Question: Are both roots positive?

Answer: Yes/No Or May be If we answer with a certain Yes or a certain No, our stmnt is sufficient. If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.

2x^2 + hx + k = 0 Sum of roots = -h/2 Product of roots = k/2

Let's find all those cases in which we can answer with 'Yes'

Stmnt 1: The product of the roots is 6. k/2 = 6 so k = 12

The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0 The roots could be no integer for any random value of h. So we have no definite answer from this statement. Not sufficient.

Stmnt 2: kh = 132 Let's see whether given this information, we can find both integer roots. kh = 132 = 2^2 * 3 * 11 If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers) So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign): k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No. k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No. So it is not possible to have both integers. We can answer with a definite 'No'.

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

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05 Nov 2016, 04:43

1

This post was BOOKMARKED

hubahuba wrote:

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.

why h & k cannot be odd? Product and sum of the roots can have decimal values.

gmatclubot

Re: If h and k are both positive integers, are both roots of the equation
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05 Nov 2016, 04:43

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